Examkin

Soru

Zorluk: OrtaPolynomials, Factor and Remainder Theorems

When the polynomial P(x)=2x3+3x2−px+qP(x) = 2x^3 + 3x^2 - px + q is divided by (x−1)(x - 1), the remainder is 33. Given that (x+2)(x + 2) is a factor of P(x)P(x), what is the value of p+qp + q?

  1. 2Cevap
  2. B
    4
  3. C
    6
  4. D
    -2

Cevap

The value of p+qp + q is 22.
According to the Remainder Theorem, P(1)=3P(1) = 3 gives 2(1)3+3(1)2−p(1)+q=32(1)^3 + 3(1)^2 - p(1) + q = 3, which simplifies to p−q=2p - q = 2. According to the Factor Theorem, (x+2)(x + 2) being a factor means P(−2)=0P(-2) = 0, giving 2(−2)3+3(−2)2−p(−2)+q=02(-2)^3 + 3(-2)^2 - p(-2) + q = 0, which simplifies to 2p+q=42p + q = 4. Solving these equations together gives p=2p = 2 and q=0q = 0. Summing them yields p+q=2p + q = 2.

Adım Adım Çözüm

1
Apply the Remainder Theorem for divisor (x−1)(x - 1)
P(1)=2(1)3+3(1)2−p(1)+q=3  ⟹  5−p+q=3  ⟹  p−q=2P(1) = 2(1)^3 + 3(1)^2 - p(1) + q = 3 \implies 5 - p + q = 3 \implies p - q = 2
By the Remainder Theorem, dividing P(x)P(x) by (x−a)(x - a) leaves a remainder equal to P(a)P(a).
2
Apply the Factor Theorem for factor (x+2)(x + 2)
P(−2)=2(−2)3+3(−2)2−p(−2)+q=0  ⟹  −16+12+2p+q=0  ⟹  2p+q=4P(-2) = 2(-2)^3 + 3(-2)^2 - p(-2) + q = 0 \implies -16 + 12 + 2p + q = 0 \implies 2p + q = 4
By the Factor Theorem, if (x−a)(x - a) is a factor of P(x)P(x), then P(a)=0P(a) = 0. Here a=−2a = -2.
3
Solve the simultaneous linear equations for pp and qq
Adding p−q=2p - q = 2 and 2p+q=42p + q = 4 yields 3p=6  ⟹  p=23p = 6 \implies p = 2. Substituting p=2p = 2 into p−q=2p - q = 2 gives q=0q = 0.
Eliminating qq allows direct calculation of pp, followed by back-substitution for qq.
4
Calculate the target expression p+qp + q
p+q=2+0=2p + q = 2 + 0 = 2
Combine the values of pp and qq to obtain the required sum.

Anahtar Kavram

Factor and Remainder Theorems for Polynomials
Bu soruyu puanla

Konular

Polynomials, Factor and Remainder Theorems

Bu konudan daha fazla soru